Python Socket Error 22 Invalid Argument
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Errno 22 Invalid Argument Python
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Oserror: [errno 22] Invalid Argument Python
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Socket Error Invalid Argument
6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up OSError [Errno 22] invalid argument when use open() in Python up vote 0 down vote favorite def choose_option(self): if self.option_picker.currentRow() == 0: description = open(":/description_files/program_description.txt","r") self.information_shower.setText(description.read()) elif self.option_picker.currentRow() == 1: requirements = open(":/description_files/requirements_for_client_data.txt", "r") self.information_shower.setText(requirements.read()) elif self.option_picker.currentRow() python setsockopt == 2: menus = open(":/description_files/menus.txt", "r") self.information_shower.setText(menus.read()) I am using resource files and something is going wrong when i am using it as argument in open function, but when i am using it for loading of pictures and icons everything is fine. python share|improve this question edited Aug 30 '14 at 15:45 CoryKramer 52.2k93983 asked Aug 30 '14 at 15:45 eugene 1111 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote That is not a valid file path. You must either use a full path open(r"C:/description_files/program_description.txt","r") Or a relative path open("program_description.txt","r") share|improve this answer answered Aug 30 '14 at 15:47 CoryKramer 52.2k93983 add a comment| up vote 0 down vote you should add one more "/" in the last "/" of path for example: open('C:\Python34\book.csv') to open('C:\Python34\book.csv') import csv with open('C:\Python34\\book.csv', newline='') as csvfile: spamreader = csv.reader(csvfile, delimiter='', quotechar='|') for row in spamreader: print(row) share|improve this answer edited Jan 5 at 18:45 Tomasz Jakub Rup 5,064714
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss so_reuseaddr the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping http://stackoverflow.com/questions/25584124/oserror-errno-22-invalid-argument-when-use-open-in-python each other. Join them; it only takes a minute: Sign up Socket.error: Invalid Argument supplied up vote 8 down vote favorite 2 I am learning networking programming and trying to grasp the basics of sockets through this example. import socket,sys s = socket.socket(socket.AF_INET,socket.SOCK_DGRAM) MAX = 65535 PORT = 1060 if sys.argv[1:] == ['server']: s.bind(('127.0.0.1',PORT)) print 'Listening at ' http://stackoverflow.com/questions/15638214/socket-error-invalid-argument-supplied , s.getsockname() while True: data,address = s.recvfrom(MAX) print ' The address at ' , address , ' says ' , repr(data) s.sendto('your data was %d bytes' % len(data),address) elif sys.argv[1:] == ['client']: print ' Address before sending ' ,s.getsockname() s.sendto('This is the message',('127.0.0.1',PORT)) print ' Address after sending ' ,s.getsockname() data,address = s.recvfrom(MAX) print ' The server at ' , address , ' says ' , repr(data) else: print >> sys.stderr, 'usage: udp_local.py server | client ' However,its throwing up an exception saying the arguments given by getsockname() were invalid specifically on line 22.The code is correct as far as I know.Here's the exception Traceback (most recent call last): File "udp_local.py", line 23, in
help? Post your question and get tips & solutions from a community of 418,626 IT Pros & Developers. It's quick & easy. getting socket.bind() exception, but no actual error P: n/a https://bytes.com/topic/python/answers/37733-getting-socket-bind-exception-but-no-actual-error Clarence Gardner I've got a problem that I don't see how to program around. A socket server that was running fine, today started getting an exception from the bind() call (errno 22, Invalid argument) and https://blenderartists.org/forum/archive/index.php/t-177473.html yet the bind had actually worked. This is apparent from this code: it prints the error message, but then breaks out of the loop and the listen() and accept() calls work normally. How can one invalid argument handle a function that is both working and raising an exception? sockobj=socket(AF_INET,SOCK_STREAM) while True: try: sockobj.bind(('',4321)) except socketerror,e: print e if str(e).find('in use') == -1: break print '.' time.sleep(1) sockobj.listen(5) Jul 18 '05 #1 Post Reply Share this Question 6 Replies P: n/a Steve Holden Clarence Gardner wrote: I've got a problem that I don't see how to program around. A socket server that was running fine, today started getting 22 invalid argument an exception from the bind() call (errno 22, Invalid argument) and yet the bind had actually worked. This is apparent from this code: it prints the error message, but then breaks out of the loop and the listen() and accept() calls work normally. How can one handle a function that is both working and raising an exception? sockobj=socket(AF_INET,SOCK_STREAM) while True: try: sockobj.bind(('',4321)) except socketerror,e: print e if str(e).find('in use') == -1: break print '.' time.sleep(1) sockobj.listen(5) Is it possible you already have a process bound to that port, and that process is handling the incoming connections? Otherwise it sounds a little bizarre. In fact even my own attempt at explaining doesn't ring true because on Win2k, at least, a second bind will raise (10048, 'Address already in use'), and I get a similar error on Linux. Rats. I take it this code has been copied and pasted? For example, if the host were a single space rather than an empty string /that/ might cause this problem. Sorry I can't be more help. regards Steve -- http://www.holdenweb.com http://pydish.holdenweb.com Holden Web LLC +1 800 494 3119 Jul 18 '05 #2 P: n/a Michael Fuhr cl******@silcom.com (Clarence Gardner) writes: And the whole point of the test in the exception handling suite
and I ran into a little problem. When I ran this script: from socket import * host = "" port = 8002 s = socket(AF_INET,SOCK_DGRAM) s.bind((host,port)) s.setblocking(0) s.listen(1) conn, addr = s.accept() print repr(conn.recv(1024)) conn.send("Hello World") print "Connected by", addr print "Main Port" ,port, "is up" I got this error from s.listen(1): Traceback (most recent call last): File "Server", line 7, in ? File "